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## Homework Statement

Let ##E## be a closed set of real numbers and ##f## a real-valued function that is defined and continuous on ##E##. Show that there exists a function ##g## defined and continuous on all of ##\Bbb{R}## such that ##f(x) = g(x)## for each ##x \in E##.

## Homework Equations

## The Attempt at a Solution

Since ##E## is closed, the set ##U = \Bbb{R} - E## must be open, which means that it can be written as a union of disjoint intervals like so: ##U = \bigcup_{i \in I} (a_i,b_i)##. Now, if ##x \in E##, simply define ##g(x) = f(x)##. However, if ##x \in U##, then there exists a ##i## such that ##x \in (a_i,b_i)##. In this case, let ##g(x) = \frac{f(b_i)-f(a_i)}{b_i-a_i}(x-a_i) + f(a_i)##. So, in other words,

$$g(x) = \begin{cases} f(x), & x \in E \\

\\

\frac{f(b_1)-f(a_1)}{b_1-a_1}(x-a_1) + f(a_1), & x \in (a_1,b_1) \\

\\

\frac{f(b_2)-f(a_2)}{b_2-a_2}(x-a_2) + f(a_2), & x \in (a_2,b_2) \\

\vdots & \vdots \\

\end{cases}$$

Now we prove continuity. To do this, we need only concern ourselves with the endpoints of the intervals. Take one of these intervals ##(a,b)##, and without loss of generality we will verify continuity at ##x = a##. I think only need to compute the right-hand limit (using the definition of course, but I will omit it since it is trivial). Since linear functions are continuous, the right-hand limit exists and equals the actual limit; i.e.,

$$\lim_{x \to a^+} g(x) = \frac{f(b)-f(a)}{b-a} \lim_{x \to a} (x-a) +f(a) = f(a),$$

which shows ##g## is continuous at ##x=a##.

What worries me is that left and right hand limits have not been introduced, and if they were suppose to be used, then they would be introduced, since Real Analysis by Royden and Fitzpatrick is a rigorous graduate level book. However, this is the only method of proof I could come up with. Maybe the book is presupposing knowledge of left and right handed limits. What do you think? Is there a better way to solve the problem?

I just realized that I left out the case in which one of the open intervals being is unbounded. That shouldn't be too hard though; I just make the sure the one finite endpoint 'matches up' with ##f##, as I did above.